设x，y，z均为正实数，且xyz=1，求证：1x3y+1y3z+1z3x≥xy+yz+zx．

Question

设x，y，z均为正实数，且xyz=1，求证：(1)/(x3y)+(1)/(y3z)+(1)/(z3x)≥xy+yz+zx．

Answer 1

证明：∵x，y，z均为正实数，且xyz=1，
∴(1)/(x3y)+(1)/(y3z)+(1)/(z3x)=(z)/(x2)+(x)/(y2)+(y)/(x2)，
∴由柯西不等式可得（(z)/(x2)+(x)/(y2)+(y)/(x2)）（xy+yz+zx）
≥（(√(xyz))/(x)+(√(xyz))/(y)+(√(xyz))/(x)）²=（(xyz)/(x)+(xyz)/(y)+(xyz)/(z)）²=（xy+yz+zx）²．
∴(1)/(x3y)+(1)/(y3z)+(1)/(z3x)≥xy+yz+zx．